# Interpreting phonon dispersion relations

I have been working with phonon dispersion relations for a while now on the topic of metamaterials (phononic band gaps). However, I still do not feel that I have fully grasped how to interpret these disperion relation diagrams.

The frequency $$\omega$$ is plotted against the wave vector $$k$$, but how do I actually read it? Do I search for a frequency and look which modes are "(co)existing" at that frequency? Or do I pick a wave vector (a direction) and look which frequencies are allowed for these values of $$k$$? I can probably read it both ways, but where is cause and effect exactly?

Here's what I know: Let's assume a 2D case with a simple Brillouin Zone $$\Gamma$$-X-Y-$$\Gamma$$. The sections of the dispersion relation correspond to values of $$k$$, where $$\Gamma$$ denotes the point where $$k$$ is very small and the wavelength $$\lambda$$ is very large. Traveling along the x-axis is basically like traversing the edges of the Brillouin Zone, covering all possible directions of the wave vector.

1. Suppose, a dispersion branch for $$\Gamma$$-X has two possible frequencies. What is the "real world meaning" of that? Do both these modes exist at a certain excitation frequency?

2. Now assume there are two different branches that occur at the same frequency inside $$\Gamma$$-X. Does that make it any different than case 1 where the same branch has one frequency twice?

## 1 Answer

I think the short version is that dispersion relations only tell a small part of the story of a phonon. They say nothing about the motion of the corresponding wave; all they tell you is how the oscillatory frequency is related to the wavevector.

As for how to read the curves: you read them just like any function. There is no inherent cause and effect, just as there is no inherent cause and effect for $$y = x^2$$. $$x=2$$ does not cause $$y=4$$ any more than $$y=4$$ causes $$x=2$$. For phonons, an oscillatory source with a given frequency would cause phonon(s) with specific wavelengths, and an excitation with a given wavelength would cause phonons(s) with specific frequencies. The cause depends on what you do --- not on a curve.

Let's start with a simpler system: an infinite, isotropic, continuum solid (e.g. jello). This systems supports two kinds of waves: longitudinal (also known as pressure or p-waves) and transverse (also known as shear or s-waves). The two types of waves will, in general, have different velocities --- say $$c_l$$ and $$c_t$$, respectively. The dispersion relation for "phonons" in this system will then have two branches: $$\omega_l = c_l \left|\vec{k}\right|$$ and $$\omega_t = c_t \left|\vec{k}\right|$$.

The real world meaning is that, at any given $$\vec{k}$$, both longitudinal and transverse waves can exist. They will, however, look quite different (see below --- taken from Wikipedia). In this system, you can find longitudinal and transverse waves to match any frequency.

When you're working with atoms (or meta materials), things become more complicated, but the long-wavelength limit of acoustic phonons converges to elastic waves in a continuum system. To answer your questions:

1. I interpret this to mean that you have a single branch with $$\omega_1\left(a\hat{x}\right) = \omega_1\left(b\hat{x}\right)$$ where $$a \neq b$$. This means that there are two excitations with different wavelengths that have the same oscillatory frequency. If you were to visualize their motion, it would be easy to distinguish them because they would have different wavelengths. (They might also look quite different for other reasons too.)
2. I interpret this to mean that you have two branches with $$\omega_1\left(a\hat{x}\right) = \omega_2\left(b\hat{x}\right)$$ where $$a \neq b$$. You can achieve this with the continuum material we started with; simply choose $$c_l a = c_t b$$. All we are now saying is that longitudinal elastic waves can have the same frequency as transverse elastic waves. Again, if you visualize the motion, you'll very easily see that the waves are different even though they oscillate with the same frequency.
• Thanks for this answer! In theory, this is all very sound (heh, it is not even sound in solids). But what does this mean practically? When I have a frequency that enables both a longitudinal and a transverse wave motion, do I get a superposition of both wave motions? If they have different wavelengths it would probably be easy to see, right?
– Ian
Commented Jul 26, 2019 at 5:42
• I can't answer an "in practice" question without knowing what you're doing. If you point a loudspeaker at a large piece of Jello and play a single frequency through the speaker, you will in general produce a combination (superposition) of longitudinal and transverse waves of the same frequency. However, if you point the speaker at some special angles, you can get just one type or the other. If you set off some explosion near the Jello (see reflection seismology, or earthquakes), you'll get some combination of transverse and acoustic waves, which can be differentiated by their speeds. Commented Jul 26, 2019 at 13:31